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Q.Prove that cos² 48° - sin² 12° = (√5 + 1)/8.

West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 2mImportance★★★★★est
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Apply the identity cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2A-\sin^2B=\cos(A+B)\cos(A-B) with A=48∘,B=12∘A=48^\circ,B=12^\circ, then use the known value cos⁡36∘=1+54\cos36^\circ=\dfrac{1+\sqrt5}{4}.

Identity used: cos⁡(A+B)cos⁡(A−B)=(cos⁡Acos⁡B−sin⁡Asin⁡B)(cos⁡Acos⁡B+sin⁡Asin⁡B)=cos⁡2Acos⁡2B−sin⁡2Asin⁡2B\cos(A+B)\cos(A-B)=(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)=\cos^2A\cos^2B-\sin^2A\sin^2B

=cos⁡2A(1−sin⁡2B)−(1−cos⁡2A)sin⁡2B=cos⁡2A−sin⁡2B.=\cos^2A(1-\sin^2B)-(1-\cos^2A)\sin^2B=\cos^2A-\sin^2B.

So cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2A-\sin^2B=\cos(A+B)\cos(A-B).

With A=48∘,B=12∘A=48^\circ, B=12^\circ: A+B=60∘A+B=60^\circ, A−B=36∘A-B=36^\circ. …

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